If \(\cos A = \frac{12}{13}\) and \(\sin B = \frac{5}{13}\), find \(\sin A \cos B + \cos A \sin B\):
\(\frac{120}{169}\)
The expression \(\sin A \cos B + \cos A \sin B\) is exactly the expansion of the compound-angle identity \(\sin(A+B)\), so we can either use that identity or evaluate each product directly after finding the missing ratios.
We are given \(\cos A = \frac{12}{13}\). Using \(\sin^2 A + \cos^2 A = 1\), we get \(\sin A = \sqrt{1 - \left(\frac{12}{13}\right)^{2}} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}\).
We are given \(\sin B = \frac{5}{13}\). Similarly, \(\cos B = \sqrt{1 - \left(\frac{5}{13}\right)^{2}} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}\).
Now compute the first product: \(\sin A \cos B = \frac{5}{13} \times \frac{12}{13} = \frac{60}{169}\).
Next the second product: \(\cos A \sin B = \frac{12}{13} \times \frac{5}{13} = \frac{60}{169}\).
Adding the two: \(\frac{60}{169} + \frac{60}{169} = \frac{120}{169}\). The key idea is the sine addition formula, which lets us read the whole expression as one clean value.
Hence the required value is \(\frac{120}{169}\).
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